The given expression is \( (x + y)^{100} + (x - y)^{100} \). To find the total number of terms in the expansion of this expression, we will first analyze the individual expansions of \( (x + y)^{100} \) and \( (x - y)^{100} \), and then combine them.
The binomial expansion of \( (x + y)^{100} \) is given by:
\[ (x + y)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^{100-k} y^k \]
This expansion has \( 101 \) terms because the index \( k \) ranges from 0 to 100, inclusive. So, the expansion of \( (x + y)^{100} \) contains 101 terms.
The binomial expansion of \( (x - y)^{100} \) is given by:
\[ (x - y)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^{100-k} (-y)^k = \sum_{k=0}^{100} \binom{100}{k} (-1)^k x^{100-k} y^k \]
Like the first expansion, this also has \( 101 \) terms. However, the signs of the terms alternate depending on whether \( k \) is even or odd due to the factor \( (-1)^k \). So, the expansion of \( (x - y)^{100} \) also contains 101 terms.
Now, consider the sum of the two expansions:
\[ (x + y)^{100} + (x - y)^{100} \] When we add these two expansions, terms where the powers of \( y \) are odd will cancel each other out, because \( (-y)^k \) will have the opposite sign of \( y^k \) when \( k \) is odd. On the other hand, terms where \( k \) is even will add up, because both expansions will have the same sign for those terms.
Thus, only the terms where \( k \) is even will remain after addition. The even values of \( k \) range from 0 to 100, inclusive. These values are \( k = 0, 2, 4, \dots, 100 \), which gives a total of 51 even values.
The total number of terms in the expansion of \( (x + y)^{100} + (x - y)^{100} \) is 51, because only the terms with even values of \( k \) survive after adding the two expansions
Let $ (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + ... + a_{20} x^{20} $. If $ (a_1 + a_3 + a_5 + ... + a_{19}) - 11a_2 = 121k $, then k is equal to _______
In the expansion of $\left( \sqrt{5} + \frac{1}{\sqrt{5}} \right)^n$, $n \in \mathbb{N}$, if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$, then the value of $^nC_3$ is:
Two point charges M and N having charges +q and -q respectively are placed at a distance apart. Force acting between them is F. If 30% of charge of N is transferred to M, then the force between the charges becomes:
If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be: